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2020-2021:teams:legal_string:数论概论学习小结_lgwza [2020/07/03 22:25] lgwza [第 25 章 二次互反律] |
2020-2021:teams:legal_string:数论概论学习小结_lgwza [2020/07/03 22:25] (当前版本) lgwza [第 25 章 二次互反律] |
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=== 定理 25.2 (广义二次互反律) === | === 定理 25.2 (广义二次互反律) === | ||
- | 设 $a$, $b$ 为正奇数, 则 $$ ()={ | + | 设 $a$, $b$ 为正奇数, 则 |
+ | $$ | ||
+ | \left(\frac{-1}{b}\right)=\left\{ | ||
+ | \begin{array}{rl} | ||
+ | 1 & if & b \equiv 1\pmod{4},\\ | ||
+ | -1 & if & b \equiv 3\pmod{4}. | ||
+ | \end{array} \right.\\ | ||
- | .\ | + | \left(\frac{2}{b}\right)=\left\{ |
+ | \begin{array}{rl} | ||
+ | 1 & if & b\equiv 1 \ or\ 7\pmod{8}, \\ | ||
+ | -1 & if & b\equiv 3\ or\ 5\pmod{8}. | ||
+ | \end{array}\right.\\ | ||
- | ()={ | + | \left(\frac{a}{b}\right)=\left\{ |
- | + | \begin{array}{rl} | |
- | .\ | + | \left(\frac{b}{a}\right) & if &a\equiv 1\pmod{4}\ or\ b\equiv1\pmod{4},\\ |
- | + | -\left(\frac{b}{a}\right)&if&a\equiv b\equiv 3\pmod{4}. | |
- | ()={ | + | \end{array}\right. |
- | + | $$ | |
- | . $$ | + |